If 0-2 mol of H2g and 2 mol of Ss are mixed in a 1 dm3 vessel at 90-C- the partial pressure of H2Sg formed according to the reaction- H2g - Ss- H2S- KP - 6-8- 10-2 would be-

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If 0.2 mol ...

Question

If $0.2 m o l$ of $H_{2} (g)$ and $2$ mol of $S (s)$ are mixed in a $1 d m^{3}$ vessel at $90^{\circ} C$ , the partial pressure of $H_{2} S (g)$ formed according to the reaction, $H_{2} (g) + S (s) ⇌ H_{2} S, K_{P} = 6.8 \times 10^{- 2}$ would be_______.

0.19 a t m

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0.38 a t m

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0.6 a t m

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0.072 a t m

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Solution

The correct option is B

0.38 a t m

Given

Moles of

H_{2}

=0.2 mol

Moles of S(s)=2 mol

Volume of vessel = 1

d m^{3}

Temperature= 273+90= 363K

K_{p} = 6.8 * 10^{- 2}

Solution

H_{2}

+ S(s) =

H_{2} S

t=0 0.2 2 0

t=t 0.2-x 2-x x

Let x be no. moles of

H^{2} S

Total no. of moles is o.2 -x +2 -x + x =2.2 - x

Mole fraction of

H_{2} S = \frac{x}{2.2 - x}

Mole fraction of

H_{2} = \frac{0.2 - x}{2.2 - x}

Kp=

\frac{[H_{2} S]}{[H_{2}]}

Kp=

\frac{x}{0.2 - x}

putting value of Kp we get x=0.0145

Partial pressure of

H_{2} S = \frac{x}{1.2 - x} * 1

=0.38atm

The correct answer is B

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Similar questions

Q. If

0.2 m o l

H_{2} (g)

and

2 m o l

S (s)

are mixed in a

1 d m^{3}

vessel at

90^{o} C,

the partial pressure of

H_{2} S (g)

formed according to the reaction would be :

H_{2} (g) + S (s) ⇌ H_{2} S; K_{P} = 6.8 \times 10^{- 2}

Q. If 0.2 mol of

H_{2} (g)

and 2.0 mol of S(s) are mixed in a 1.0 litre vessel at

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, the partial pressure of

H_{2} S (g)

formed according to the reaction,

H_{2} (g) + S (s) ⟺ H_{2} S (g)

;

K_{p} = 6.8 \times 10^{- 2}

would be :

Q. If

0.2 m o l

H_{2} (g)

and

2 m o l

S (s)

are mixed in a

1 d m^{3}

vessel at

90^{o} C,

the partial pressure of

H_{2} S (g)

formed according to the reaction would be :

H_{2} (g) + S (s) ⇌ H_{2} S; K_{P} = 6.8 \times 10^{- 2}

Q. At

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, the following equilibrium is established

H_{2} (g) + S (s) ⇌ H_{2} S (g)

K_{p} = 6.8 \times 10^{- 2}

0.2

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1.0

mol of sulphur are heated to

90^{o} C

in a

1.0

litre vessel, what will be the partial pressure of

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Q. At

400^{\circ} C

, the following equilibrium is established :

H_{2} (g) + S (s) ⇌ H_{2} S (g)

K_{p} = 6.8 \times 10^{- 2}

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If 0.2 mol of H2(g) and 2 mol of S(s) are mixed in a 1 dm3 vessel at 90∘C, the partial pressure of H2S(g) formed according to the reaction, H2(g)+S(s)⇌H2S,KP=6.8×10−2 would be_______.

If $0.2 m o l$ of $H_{2} (g)$ and $2$ mol of $S (s)$ are mixed in a $1 d m^{3}$ vessel at $90^{\circ} C$ , the partial pressure of $H_{2} S (g)$ formed according to the reaction, $H_{2} (g) + S (s) ⇌ H_{2} S, K_{P} = 6.8 \times 10^{- 2}$ would be_______.